# Maths question

Homework Helper
Hi,
I have been trying to solve it for some days,... but I couldnot find out. The equation is:
$$a^{2} + b^{2} = c^{2}$$
a, b, c <> 0
I try to call:
$$x_{1} = \frac{b}{a}$$
$$x_{2} = \frac{c}{a}$$
And I have
$$a^2 \times{(1 + x_{1}^{2} - x_{2}^{2})} = 0$$
And I solve for x1 and x2. But I didn't succeed.
Can you suggest me some other ways??
Thanks,
Viet Dao,

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arildno
Homework Helper
Gold Member
Dearly Missed
What are you after?
Is it a formula for how pythogorean triplets in general can be found?
Be more specific!

dextercioby
Homework Helper
If this must be the case,then:

$$a=m^{2}-n^{2}$$

$$b=2mn$$

$$c=m^{2}+n^{2}$$

Where "m" and "n" are ARBITRARY.

Daniel.

Homework Helper
arildno said:
What are you after?
Is it a formula for how pythogorean triplets in general can be found?
Be more specific!
I dunno really. My friend just asked me that some weeks ago, and I simply got stuck.
I remember reading a book that says:
$$a^{n} + b^{n} = c^{n}$$
Where a, b, c <> 0 and $a, b, c \in Z$
$n \in N$, n > 2. Then there will be no answer. But it didn't prove it.
--------------
Thanks Dextercioby, I got it now... But how can you know that?
$$a=m^{2}-n^{2}$$
$$b=2mn$$
$$c=m^{2}+n^{2}$$
Is there another case?
--------------
Viet Dao,

dextercioby
Homework Helper
I didn't invent it,certainly...I read it somewhere,a long time ago.And it's not too hard to remember.

No,that formula covers all possible "m" and "n" from R (in particular from N hence generating the pythagoreic triplets).

Daniel.

Homework Helper
Hi,
Thanks a lot dextercioby. That really helps.
Viet Dao,

Hurkyl
Staff Emeritus
Gold Member
I think you need an additional factor to get all triples:

a = p(m2 - n2)
b = 2mnp
c = p(m2 + n2)

Of course, you don't need p for primitive triples.

dextercioby
Homework Helper
I see what u mean.Yes,the triplet (6,12,15) would not be found through "m" and "n"...

Daniel.

Homework Helper
Hi,
Thanks, Hurkyl. I appreciate your help.
Viet Dao,

dextercioby